A363674 -id:A363674 - cp's OEIS frontend

A329278 Irregular table read by rows. The n-th row is the permutation of {0, 1, 2, ..., 2^n-1} given by T(n,k) = k(k+1)/2 (mod 2^n).

0, 0, 1, 0, 1, 3, 2, 0, 1, 3, 6, 2, 7, 5, 4, 0, 1, 3, 6, 10, 15, 5, 12, 4, 13, 7, 2, 14, 11, 9, 8, 0, 1, 3, 6, 10, 15, 21, 28, 4, 13, 23, 2, 14, 27, 9, 24, 8, 25, 11, 30, 18, 7, 29, 20, 12, 5, 31, 26, 22, 19, 17, 16, 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 2

Offset: 0

Peter Kagey, Nov 11 2019

Conjecture: for n > 0, the n-th row has 2^(n-1)-1 descents.

T(n,k) = A000217(k) for 0 <= k <= A017911(n+1), and T(n,2^n-1) = 2^(n-1).

			Table begins:
  0;
  0, 1;
  0, 1, 3, 2;
  0, 1, 3, 6,  2,  7, 5,  4;
  0, 1, 3, 6, 10, 15, 5, 12, 4, 13, 7, 2, 14, 11, 9, 8;
  ...

Peter Kagey, Table of n, a(n) for n = 0..8190 (first 12 rows)

Cf. A000217, A000225, A017911, A053645, A105332, A330766, A331105, A363674.

T(n,2^n-1) gives A131577.

T:= (n, k)-> irem(k*(k+1)/2, 2^n):
seq(seq(T(n, k), k=0..2^n-1), n=0..6);  # Alois P. Heinz, Jan 08 2020

A331105 T(n,k) = -k*(k+1)/2 mod 2^n; triangle T(n,k), n>=0, 0<=k<=2^n-1, read by rows.

Original entry on oeis.org

0, 0, 1, 0, 3, 1, 2, 0, 7, 5, 2, 6, 1, 3, 4, 0, 15, 13, 10, 6, 1, 11, 4, 12, 3, 9, 14, 2, 5, 7, 8, 0, 31, 29, 26, 22, 17, 11, 4, 28, 19, 9, 30, 18, 5, 23, 8, 24, 7, 21, 2, 14, 25, 3, 12, 20, 27, 1, 6, 10, 13, 15, 16, 0, 63, 61, 58, 54, 49, 43, 36, 28, 19, 9

Offset: 0

Alois P. Heinz, Jan 09 2020

Row n is a permutation of {0, 1, ..., A000225(n)}.

			Triangle T(n,k) begins:
  0;
  0,  1;
  0,  3,  1,  2;
  0,  7,  5,  2, 6, 1,  3, 4;
  0, 15, 13, 10, 6, 1, 11, 4, 12, 3, 9, 14, 2, 5, 7, 8;
  ...

Alois P. Heinz, Rows n = 0..15, flattened

Columns k=0-2 give: A000004, A000225, A036563 (for n>1).
Row sums give A006516.
Row lengths give A000079.
T(n,n) gives A014833 (for n>0).
T(n,2^n-1) gives A131577.
Cf. A329278, A363674.

T:= n-> (p-> seq(modp(-k*(k+1)/2, p), k=0..p-1))(2^n):
seq(T(n), n=0..6);
# second Maple program:
T:= proc(n, k) option remember;
      `if`(k=0, 0, T(n, k-1)-k mod 2^n)
    end:
seq(seq(T(n, k), k=0..2^n-1), n=0..6);

T[n_, k_] := T[n, k] = If[k == 0, 0, Mod[T[n, k - 1] - k, 2^n]];
Table[Table[T[n, k], {k, 0, 2^n - 1}], {n, 0, 6}] // Flatten (* Jean-François Alcover, Mar 28 2022, after Alois P. Heinz *)

A362160 Irregular triangle read by rows: The n-th row contains 2^n integers corresponding to the words of the n-bit Gray code with the most significant bits changing fastest.

Original entry on oeis.org

0, 0, 1, 0, 2, 3, 1, 0, 4, 6, 2, 3, 7, 5, 1, 0, 8, 12, 4, 6, 14, 10, 2, 3, 11, 15, 7, 5, 13, 9, 1, 0, 16, 24, 8, 12, 28, 20, 4, 6, 22, 30, 14, 10, 26, 18, 2, 3, 19, 27, 11, 15, 31, 23, 7, 5, 21, 29, 13, 9, 25, 17, 1, 0, 32, 48, 16, 24, 56, 40, 8, 12, 44, 60, 28

Offset: 0

Valentin Bakoev, Apr 10 2023

The n-th row of the triangle is defined recursively as row(0) = 0 which corresponds to the empty word, and row(n) = row(n-1)0, row^r(n-1)1, for n > 0. Here row(n-1)0 is the sequence of words of the (n-1)-bit Gray code of this type suffixed with 0, and row^r(n-1)1 means the sequence of reflected words (i.e., words are taken in reverse order) of the (n-1)-bit Gray code of this type and then each word is suffixed with 1.
Another way to obtain row(n) is by applying the transition sequence A001511(n), which indicates which bit to flip in the current word to get the next word - see the FORMULA section.
If we reverse the internal order of bits in each word of row(n), we obtain the binary reflected n-bit Gray code (see A003188) and vice versa.

			Triangle begins:
    k = 0   1   2  3   4   5   6  7 ...
  n=0:  0,
  n=1:  0,  1,
  n=2:  0,  2,  3, 1,
  n=3:  0,  4,  6, 2,  3,  7,  5, 1,
  n=4:  0,  8, 12, 4,  6, 14, 10, 2, 3, 11, 15, 7, 5, 13, 9, 1,
  n=5:  0, 16, 24, 8, 12, 28, 20, 4, 6, 22, 30, 14, 10, 26, 18, 2, 3, 19, 27, 11, 15, 31, 23, 7, 5, 21, 29, 13, 9, 25, 17, 1,
  ...
In row n=3, the corresponding binary words of length 3 are 000, 100, 110, 010, 011, 111, 101, and 001 - notice that the most significant bits change the fastest.

W. Lipski Jr, Combinatorics for programmers, Mir, Moscow, 1988, (in Russian), p. 31, Algorithm 1.13.
F. Ruskey, Combinatorial Generation. Working Version (1j-CSC 425/520), 2003, pp. 120-121.

Valentin Bakoev, Rows n = 0..15, flattened
Valentin Bakoev, Mirror (left-recursive) Gray Code, Mathematics and Informatics, Vol. 66, Number 6, pp. 559-578, (2023).

Cf. A003188 (Gray code), A006516 (row sums), A000004 (column k=0), A000079 (column k=1), A088208.
T(n,2^n-1) gives A057427.
Cf. also A000120, A005811, A363674.

with(ListTools): with(Bits):
T:= (n, k)-> Join(Reverse(Split(Xor(k, iquo(k, 2)), bits=n))):
seq(seq(T(n, k), k=0..2^n-1), n=0..6);  # Alois P. Heinz, Jun 05 2023

T(n,k) = fromdigits(Vecrev(binary(bitxor(k,k>>1)), n), 2); \\ Kevin Ryde, Apr 17 2023

T(n,k) = 2*T(n-1,k) for 0 <= k < 2^(n-1), and
T(n,2^(n-1)+k) = 2*T(n-1,2^(n-1)-k-1) + 1 = T(n,2^(n-1)-k-1) + 1 for 0 <= k < 2^(n-1).
T(n,k+1) = T(n,k) XOR 2^(n-A001511(k)).
A000120(T(n,n)) = A005811(n). - Alois P. Heinz, Jun 27 2023
T(n, k) = A088208(n, k) - 1. - Andrey Zabolotskiy, Dec 06 2024

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A329278 Irregular table read by rows. The n-th row is the permutation of {0, 1, 2, ..., 2^n-1} given by T(n,k) = k(k+1)/2 (mod 2^n).

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A331105 T(n,k) = -k*(k+1)/2 mod 2^n; triangle T(n,k), n>=0, 0<=k<=2^n-1, read by rows.

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A362160 Irregular triangle read by rows: The n-th row contains 2^n integers corresponding to the words of the n-bit Gray code with the most significant bits changing fastest.

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